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Concepts and Vocabulary
True or False The directional derivative Duf(x0,y0) equals the rate of change of z=f(x,y) at (x0,y0) in the direction of the unit vector u.
True
True or False For a function z=f(x,y), the partial derivative fx(x0,y0) is the directional derivative at (x0,y0) in the direction of i.
True
True or False Both the directional derivative Duf(x,y) and the gradient ∇f(x,y) of a function z=f(x,y) are vectors.
False
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True or False If z=f(x,y) is a differentiable function, then the directional derivative of f at (x0,y0) in the direction of u=cosθi+sinθj is given by Duf(x0,y0)=fx(x0,y0)cosθ+fy(x0,y0)sinθ.
True
True or False The vector xi+sinyj is the gradient of f(x,y)=xsiny.
False
The directional derivative Duf(x0,y0) of f at (x0,y0) is a maximum when u=_____.
1‖
The value of z=f(x,y) at (x_{0},y_{0}) decreases most rapidly in the direction of _____.
-\nabla f(x_0,y_0)
For a differentiable function z=f(x,y), the maximum value of D_{\mathbf{u}}f(x_{0},y_{0}) is _____.
\left\Vert{\nabla f(x_0 ,y_0 )} \right\Vert
Skill Building
In Problems 9–16,
f(x,y)=xy^{2}+x^{2} at (-1,2) in the direction \mathbf{u}= \dfrac{1}{2}\mathbf{i}+\dfrac{\sqrt{3}}{2}\mathbf{j}
f(x,y)=3xy+y^{2} at (2,1) in the direction \mathbf{u}= \dfrac{\sqrt{2}}{2}\mathbf{i}+\dfrac{\sqrt{2}}{2}\mathbf{j}
f(x,y)=2xy-y^{2} at (-1,3) in the direction \theta =\dfrac{2\pi }{3}
f(x,y)=2xy+x^{2} at (0,3) in the direction \theta = \dfrac{4\pi }{3}
f(x,y)=xe^{y}+ye^{x} at (0,0) in the direction \theta = \dfrac{\pi }{6}
f(x,y)=x\ln y at (5,1) in the direction \theta =\dfrac{ \pi }{4}
f(x,y)=\tan ^{-1}\dfrac{y}{x} at (1,1) in the direction of \mathbf{a}=3\mathbf{i}-4\mathbf{j}
f(x,y)=\ln \sqrt{x^{2}+y^{2}} at (3,4) in the direction of \mathbf{a}=5\mathbf{i}+12\mathbf{j}
In Problems 17–32:
f(x,y)=xy^{2}+x^{2};\qquad P=(1,2), Q=(2,4)
f(x,y)=2xy+x^{2};\qquad P=(-1,1), Q=(1,2)
f(x,y)=2xy+x^{2};\qquad P=(0,3), Q=(4,1)
f(x,y)=3xy+y^{2};\qquad P=(2,1), Q=(4,1)
f(x,y)=xy+\sin x;\qquad P=(0,1), Q=(\pi ,2)
f(x,y)=e^{xy}+\sin y;\qquad P=\left( 0,\dfrac{\pi }{2}\right), Q=(1,0)
f(x,y)=\tan ^{-1}\dfrac{y}{x};\qquad P=(1,0), Q=(4,\pi )
f(x,y)=\ln \sqrt{x^{2}+y^{2}};\qquad P=(3,4), Q=(0,5)
f(x,y)=x^{2}e^{y};\qquad P=(2,0), Q=(3,0)
f(x,y)=e^{x^{2}+y^{2}};\qquad P=(1,2), Q=(2,3)
f(x,y,z)=x^{2}y-xyz^{2};\qquad P=(0,1,2), Q=(1,4,3)
f(x,y,z)=x^{2}y+y^{2}z+z^{2}x;\qquad P=(1,2,-1), Q=(2,0,1)
f(x,y,z)=z\tan ^{-1}\dfrac{y}{x};\qquad P= (1,1,3), Q= ( 2,0,-2)
f(x,y,z)=\sin x\cos (y+z);\qquad P= (1,1,1), Q= (2,-1,0)
f(x,y,z)=\sqrt{x^{2}+y^{2}+z^{2}};\qquad P= (3,4,0), Q= ( 1,-1,1)
f(x,y,z)=\dfrac{x}{\sqrt{x^{2}+2y^{2}+3z^{2}}};\qquad P= (1,2,1), Q= ( -1,1,1)
In Problems 33–40:
f(x.y) =xy^{2}+x^{2} at P=(-1,2)
f(x.y) =3xy+y^{2} at P=(2,1)
f(x.y) =xe^{y}+ye^{x} at P=(0,0)
f(x.y) =x\ln y at P=(5,1)
f(x,y)=\dfrac{x}{x^{2}+y^{2}} at P=(1,2)
f(x,y)=\sqrt{x^{2}+y^{2}} at P=(3,4)
f( x.y,z) =z\tan ^{-1}\dfrac{y}{x} at P=(1,1,3)
f( x.y,z) =\sqrt{x^{2}+y^{2}+z^{2}} at P=(3,4,0)
In Problems 41–46, graph the level curve of f containing the point P and the gradient {\bf\nabla\! }f at that point.
f(x,y)=x^{2}+y^{2} containing P=(3,4)
f(x,y)=x^{2}-y^{2} containing P=(2,-1)
f(x,y)=x^{2}-4y^{2} containing P=\left( 3,\dfrac{\sqrt{5}}{2}\right)
f(x,y)=x^{2}+4y^{2} containing P=(-2,0)
f(x,y)=x^{2}y containing P=\left( 3,\dfrac{1}{9}\right)
f(x,y)=xy containing P=(1,1)
Applications and Extensions
Heat Transfer A metal plate is placed on the xy-plane in such a way that the temperature T at any point (x,y) is given by T=e^{x}( \sin x+\sin y) {}^{\circ}{\rm C}.
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Electrical Potential The electrical potential V at any point (x,y) is given by V=\ln \sqrt{x^{2}+y^{2}}. Find the rate of change of potential V at any point (x,y)\neq (0,0):
Geography The surface of a hill is modeled by the equation z=( 8-2x^{2}-y^{2}) \rm{m}. See the figure below. If a freshwater spring is located at the point (1,2,2), in what direction will the water flow?
The graph of z and its level curves.
Water will flow in the direction -\nabla{\bf z}(1,2)=4{\bf i}+4{\bf j}.
Temperature Change Suppose that the temperature in degrees Celsius at each point of the coordinate plane is T=( 3x^{2}+4y^{2}+5) {}^{\circ}{\rm C}.
Temperature Change The temperature at any point (x,y) of a rectangular plate lying in the xy-plane is given by T=[ x\sin ( 2y) ] {}^{\circ}{\rm C}. Find the rate of change of temperature at the point \left( 1,\dfrac{\pi }{4}\right) in the direction making an angle of \dfrac{\pi }{6} with the positive x-axis.
The rate of change of temperature is {{\sqrt 3 } \over 2} ^{\circ}C per unit of distance in the u-direction.
Rate of Change Suppose that z=xy^{2}. In what direction(s) can we go from the point (-1,1) if we want the rate of change of z to be 2?
Rate of Change
The graph of z and its level curves. The curve 4x^{2} +9y^{2} =25 appears in red.
Show that the level curves of z = f(x,y) = x^{2}-y^{2} are orthogonal to the level curves of h(x,y)=xy for all (x_{0},y_{0})\neq (0,0). See the figure below.
The level curves of z are in blue. The level curves of h are in red.
Find a unit vector \mathbf{u} that is normal to the level curve of f(x,y)=4x^{2}y through P=(1,-2) at P. See the figure below.
The level curves of f(x,y) = 4x^{2}y.
Vector is normal to the level curve of f at P, as is
Find a unit vector \mathbf{u} that is normal to the level curve of f(x,y)=2x^{2}+y^{2}+1 through P=(1,1) at P. See the figure below.
The level curves of f(x,y) = 2x^{2}+y^{2}+1.
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Isothermal Curves Using the information from Example 5:
Mountaineering Suppose that you are climbing a mountain whose surface is modeled by the equation x^{2}+y^{2}-5x+z=0. (The x-axis points east, the y-axis north, and the z-axis up; units are in thousands of feet.)
The graph of z and its level curves.
Electrical Potential Suppose that the electrical potential (voltage V) at each point in space is V=e^{xyz} volts and that electric charges move in the direction of greatest potential drop (most rapid decrease of potential). In what direction does a charge at the point (1,-1,2) move? How fast does the potential change as the charge leaves this point?
The charge moves in the direction {\bf u} = {2 \over 3}{\bf i} - {2 \over 3}{\bf j} + {1 \over 3}{\bf k}. The electrical potential changes at a rate of -3e^{-2} volts per unit change in the u-direction.
Electric Field The electric field vector \mathbf{E} is the negative of the gradient of the electrostatic potential V. That is, \mathbf{E}=-{\bf\nabla}V. If Q is a point charge at the origin, then the electrostatic potential V at the point P(x,y,z) is given by V= \dfrac{kQ}{\sqrt{x^{2}+y^{2}+z^{2}}}, where k is a constant.
Modeling Temperature Change The temperature at each point of the region x^{2}+y^{2}+z^{2}\leq 9 is T=\sqrt{9-x^{2}-y^{2}-z^{2}} {}^{\circ}{\rm C}. If we start at the point (0,1,2) and move across the region in a straight path ending at (2,1,2), find the rate of change of T at an arbitrary point on the path.
-\dfrac{x}{\sqrt{4-x^2}}
Chemotaxis Chemotaxis is the phenomenon in which bodily cells, bacteria, and other organisms direct their movement according to a chemical stimulus. Suppose the concentration of a certain biochemical distribution at a wounded place is given by w=f(x,y,z)=\dfrac{1}{x^2 + y^2 +z^2}. If a cell that reacts by chemotaxis is located at the point (1,2,3), find the direction it will move by chemotaxis if the direction of movements is along the gradient of f.
Source: Problem submitted by the students at Minnesota State University
Gravitational Potential Energy In the study of mechanics in physics, it is shown that the gravitational force \mathbf{F}_{g} on an object equals the negative of the gradient of the gravitational potential energy U_{g} of the object. That is, \mathbf{F}_{g}=-{\bf\nabla } U_{g}. In a uniform gravitational field (such as near Earth’s surface), the gravitational potential energy is given by U_{g}=mgz, where m is the mass of an object, z (in meters) is its altitude above Earth’s surface, and g=9.8\; \rm{m}/\rm{s}^{2}. Use the gradient {\bf\nabla }U_{g} to show that the gravitational force is downward and has magnitude mg.
See Student Solutions Manual.
Gravitational Field Two objects of masses M and m are a distance r, in meters, apart. The gravitational potential energy between these objects is given by U_g=-G\dfrac{mM}{r}, where G=6.67\times 10^{-11}\;\rm{Nm}^{2}/\rm{kg}^{2}. If one object is at the origin and the other at the point (x,y,z), then the gravitational field \mathbf{F}_{g} equals -\bf\nabla U_g. Show that the magnitude of the force \mathbf{F}_{g} equals G\dfrac{mM}{r^2}.
Source: Problem contributed by the students at Minnesota State University
If the function z=f(x,y) is differentiable at a point P_{0}=(x_{0},y_{0}) and {\bf\nabla\! }f(x_{0},y_{0})\neq \mathbf{0}, show that D_{\mathbf{u}}f(x_{0},y_{0})=0 in the direction orthogonal to that of {\bf\nabla\! }f(x_{0},y_{0}).
See Student Solutions Manual.
Algebraic Properties of the Gradient If u=f(x,y) and v=g(x,y) are differentiable, show that:
Show that for a nonzero vector \mathbf{a}=a_{1}\mathbf{i} +a_{2}\mathbf{j} and a differentiable function z=f(x,y), D_{\mathbf{u}}f(x,y)=\frac{a_{1}\dfrac{\partial f}{\partial x}+a_{2}\dfrac{ \partial f}{\partial y}}{\sqrt{a_{1}^{2}+a_{2}^{2}}}
where \mathbf{u}=\dfrac{\mathbf{a}}{\Vert \mathbf{a}\Vert }.
See Student Solutions Manual.
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Let F(x,y)=0 be the equation of a curve in the xy-plane. If F is differentiable and if (x_{0},y_{0}) is a point on the curve, show eak that {\bf\nabla\! }F(x_{0},y_{0}) is normal to the curve at (x_{0},y_{0}).
Use Problem 68 to show that the tangent line to the curve F(x,y)=0 at (x_{0},y_{0}) is given by a(x-x_{0})+b(y-y_{0})=0, where a=F_{x}(x_{0},y_{0}) and b=F_{y}(x_{0},y_{0}). (Assume that a and b are not both 0.)
See Student Solutions Manual.
Use Problem 69 to find the following tangent lines to the hyperbola x^{2}-y^{2}=16:
Suppose z=f(x,y) has directional derivatives in all directions. Must f be differentiable? Explain.
Function f is not necessarily differentiable. See Student Solutions Manual.
If f(x,y,z)=z^{3}+3xz-y^{2}, find the directional derivative of f at (1,2,1) in the direction of the line x-1=y-2=z-1.
Challenge Problems
Let \mathbf{r}=x\mathbf{i}+y\mathbf{j} and r=\left\Vert \mathbf{r}\right\Vert =\sqrt{x^{2}+y^{2}}.
See Student Solutions Manual.
Find the directional derivative of f(x,y)=\left\{ \begin{array}{@{}c@{\quad}c@{\quad}c} \dfrac{\sin (x^{2}+y^{2})}{x^{2}+y^{2}} & \hbox{if} & (x,y)\neq (0,0) \\ 1 & \hbox{if} & (x,y)=(0,0) \end{array} \right.
at (0,0) in the direction of \mathbf{a}=\mathbf{i}+\mathbf{j}.
Show that if {\bf\nabla\! }f(x,y)=c(x\mathbf{i}+y\mathbf{j}) , where c is a constant, then f(x,y) is constant on any circle of radius k, centered at (0,0).
See Student Solutions Manual.
The Gradient Is Normal to the Level Surface Show that if the function w=f(x,y,z) is differentiable at a point P_{0}=(x_{0},y_{0},z_{0}), and if {\bf\nabla\! }f(x_{0},y_{0},z_{0}) \neq \mathbf{0}, then the gradient {\bf\nabla\! }f(x_{0},y_{0},z_{0}) is normal to the level surface of f through P_{0}.
Let z=f(x,y) have continuous second-order partial derivatives. If \mathbf{u}=u_{1}\mathbf{i}+u_{2}\mathbf{j} is a unit vector, we have a directional derivative D_{\mathbf{u}}f(x,y)=g(x,y). If g(x,y) is differentiable and \mathbf{v}=v_{1}\mathbf{i}+v_{2}\mathbf{j} is a unit vector, we have a directional derivative D_{\mathbf{v}}g(x,y). We can view this second quantity as a second-order directional derivative for z=f(x,y). Express it in terms of f_{xx}, f_{xy}, and f_{yy} by showing that it has the value u_{1}v_{1}f_{xx}+(u_{1}v_{2}+u_{2}v_{1})f_{xy}+u_{2}v_{2}f_{yy}
See Student Solutions Manual.
Assuming that b\neq 0 in Problem 69, show that the slope of the tangent line to the curve F(x,y)=0 at (x_{0},y_{0}) is m=-\dfrac{ F_{x}(x_{0},y_{0})}{F_{y}(x_{0},y_{0})}. (This is a proof of the Implicit Differentiation Formula I from Chapter 12, Section 12.5.)